halorunner115
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The force required to stretch a Hooke’s-law
spring varies from 0 N to 62.4 N as we stretch
the spring by moving one end 17.2 cm from
its unstressed position.
Find the force constant of the spring.
Answer in units of N/m.
Find the work done in stretching the spring.
Answer in units of J.

Respuesta :

strahler19

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K


Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J


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